Problem: $\overline{AC}$ is $10$ units long $\overline{BC}$ is $5$ units long $\overline{AB}$ is $5\sqrt{5}$ units long What is $\sin(\angle ABC)$ ? $A$ $C$ $B$ $10$ $5$ $5\sqrt{5}$
Solution: SOH CAH TOA in = pposite over ypotenuse opposite $= \overline{AC} = 10$ hypotenuse $= \overline{AB} = 5\sqrt{5}$ $\sin(\angle ABC)=\frac{10}{5\sqrt{5}}$ $=\dfrac{2\sqrt{5} }{5}$